The graph of f(x) = a/x + q is given below. A is the point (-4; 3) and B is the point (2; 6) while C is (x; y) which is the x-intercept, q is asymptotes.

$$3 = \frac{a}{-4} + q \quad \Rightarrow \quad 12 = -a + 4q \quad \text{(Equation 1)}$$

Substitute point B $(2, 6)$ into the equation.

$$6 = \frac{a}{2} + q \quad \Rightarrow \quad 12 = a + 2q \quad \text{(Equation 2)}$$

Add Equation 1 and Equation 2 to eliminate $a$.

$$(12) + (12) = (-a + 4q) + (a + 2q)$$ $$24 = 6q$$ $$q = \frac{24}{6}$$ $$q = 4$$

Substitute $q=4$ into Equation 2 to find $a$.

$$12 = a + 2(4)$$ $$12 = a + 8$$ $$a = 12 - 8$$ $$a = 4$$

The values are $a=4$ and $q=4$.

$$\boxed{a=4, q=4}$$

1.1.2 Show that the equation of $f$ is $f(x) = \frac{4}{x} + 4$.

$$f(x) = \frac{4}{x} + 4$$

This matches the given equation.

1.1.3 The range of $f$.

For a hyperbola of the form $f(x) = \frac{a}{x} + q$, the horizontal asymptote is $y=q$. The range includes all real numbers except the value of the horizontal asymptote.
Since $q=4$, the horizontal asymptote is $y=4$.
The range of $f$ is all real numbers except $y=4$.

$$\boxed{y \in \mathbb{R}, y \neq 4}$$

1.1.4 The domain of $f$.

Therefore, $x \neq 0$.
The domain of $f$ is all real numbers except $x=0$.

$$\boxed{x \in \mathbb{R}, x \neq 0}$$

1.1.5 The coordinates of C.

Set the equation $f(x) = \frac{4}{x} + 4$ to 0.

$$0 = \frac{4}{x} + 4$$ $$-4 = \frac{4}{x}$$

Multiply both sides by $x$.

$$-4x = 4$$ $$x = \frac{4}{-4}$$ $$x = -1$$

The coordinates of C are $(-1, 0)$.

$$\boxed{C(-1, 0)}$$